The Mean Value Theorem. Verify that the function

Question

The Mean Value Theorem. Verify that the function F(X) = 5x^4 - 8x^3 - 22x^2 + 15x +6 satisfies the hypothesis of the Mean Value Theorem on the interval [-1, 3] and find all numbers c that satisfy the conclusion of the Mean Value Theorem.

Answer 1

First F(x) must be continuous on the interval [-1,3] and it is.

F(x) is also differentiable on the interval (-1,3)

F(-1) = 5 -8(-1)-22+15(-1)+6 = -18

F(3) = 4(81)-8(27)-22(9)+15(3)+6 = 42

F'(x) = 20x³-24x²-44x+15

F'(c) =(F(b)-F(a))/(b-a) = (42-(-18))/(3-(-1)) = 60/4 = 15

20c³-24c²-44c +15 -15 = 0

c(20c²-24c-44) = 0

c = 0, 2.2, -1

But -1 is not in (-1,3), so c = 0,2.2

Answer 2

First, find (f(3) - f(-1))/(3-(-1)) = 15

Then, f'(x) = 15 for some values of x.

Set f'(x) = 15 and solve for x. Subtracting 15 from each side gives you 20x^3 - 24x^2 - 44x = 0

It's easy to solve after it's in this form

Answer 3

f(-1) = - 18; f(3) = 42; Polinomial function satisfy conditions of Mean Value Theorem; f'(x) = 20x³ - 24x² - 44x + 15; f'(c) = (f(3) - f(-1))/(3 - (- 1)) = (42 + 18)/(3 + 1) = 15; 20c³ - 24c² - 44c + 15 = 15; 4c(5c² - 6c - 11) = 0;

c = 0; 5c² - 6c - 11 = 0; c = (3 ±√9 + 55)/5 = (3 ± 8)/5; c = 11/5 or c = - 1. But c = - 1 is end point of interval

So, we have c =0 and c = 11/5 -